What is the extraneous solution to these equations? $\dfrac{x^2 + 6}{x - 5} = \dfrac{-5x}{x - 5}$
Solution: Multiply both sides by $x - 5$ $ \dfrac{x^2 + 6}{x - 5} (x - 5) = \dfrac{-5x}{x - 5} (x - 5)$ $ x^2 + 6 = -5x$ Subtract $-5x$ from both sides: $ x^2 + 6 - (-5x) = -5x - (-5x)$ $ x^2 + 6 + 5x = 0$ Factor the expression: $ (x + 2)(x + 3) = 0$ Therefore $x = -2$ or $x = -3$ The original expression is defined at $x = -2$ and $x = -3$, so there are no extraneous solutions.